---- > [!definition] Definition. ([[matrix of a bilinear form]]) > Let $V$ be an $n$-dimensional $k$-[[vector space]] with [[basis]] $\{ \boldsymbol e_{1},\dots,\boldsymbol e_{n} \}$. Let $B=\langle \cdot, \cdot \rangle:V \times V \to k$ be a [[bilinear map|bilinear form]] on $V$. > The matrix $\boldsymbol A \in {k}^{n \times n}$ defined by $A_{ij}=\langle \boldsymbol e_{i}, \boldsymbol e_{j} \rangle$ is called the **matrix of $B$ on the basis $\{ \boldsymbol e_{1}, \dots, \boldsymbol e_{n} \}$**. Such a definition makes sense because we have $\langle \boldsymbol x, \boldsymbol y \rangle = \boldsymbol x^{\top} \boldsymbol A \boldsymbol y =\sum_{i,j=1}^{n} x_{i}A_{ij}y_{j}. $ Note that $\boldsymbol A$ is in fact the [[matrix]] of the [[linear map]] $\begin{align}V &\to V^{*} \\v & \mapsto \langle v, - \rangle \end{align}$ with respect to the basis $\{ \boldsymbol e_{i} \}_{i=1}^{n}$ and its [[dual basis]] $\{ \varphi_{i} \}_{i=1}^{n}$.[^1] ^definition Note that $\boldsymbol A$ is a [[Gram matrix]] $\boldsymbol E'\boldsymbol E$ [^1]: To see this, we need to find how to write $\flat(\boldsymbol e_{i})=\langle \boldsymbol e_{i}, - \rangle$ as a [[linear combination]] of the $\varphi_{j}$ for each $i$. We have$\begin{align} \langle \boldsymbol e_{i}, \boldsymbol v \rangle & = \left\langle \boldsymbol e_{i}, \sum_{j} v_{j} \boldsymbol e_{j} \right\rangle \\ = & \sum_{i} v_{j}\langle \boldsymbol e_{i}, \boldsymbol e_{j} \rangle \\ = & \sum_{i} \langle \boldsymbol e_{i}, \boldsymbol e_{j} \rangle \varphi_{j}(\boldsymbol v), \end{align} $ that is, the required coefficients are $\{ \langle \boldsymbol e_{i} , \boldsymbol e_{j}\rangle \}_{j=1}^{n}$ as claimed. > [!justification] > Using bilinearity: > $\begin{align} > \langle \boldsymbol x, \boldsymbol y \rangle = & \left\langle \sum_{i=1}^{n} x_{i} \boldsymbol e_{i} , \sum_{j=1}^{n} y_{j} \boldsymbol e_{j} \right\rangle \\ > = & \sum_{i=1}^{n} x_{i} \left\langle \boldsymbol e_{i}, \sum_{i=1} y_{i} \boldsymbol e_{i} \right\rangle \\ > = & \sum_{i=1} ^{n} x_{i} \sum_{j=1}^{n} y_{i} \langle \boldsymbol e_{i}, \boldsymbol e_{j} \rangle \\ > = & \sum_{i,j=1}^{n} x_{i} y_{i}A_{ij}. > \end{align}$ > And this equals $\boldsymbol x^{\top} \boldsymbol A \boldsymbol y$ because > $\begin{align} > \boldsymbol x^{\top} \boldsymbol A \boldsymbol y = & \boldsymbol x^{\top} \left( \overbrace{ \sum_{j=1}^{n} \boldsymbol A_{:,j}y_{j}}^{:=\boldsymbol z} \right) \\ > = & \boldsymbol x^{\top} \boldsymbol z \\ > = & \sum_{i=1}^{n} x_{i} z_{i} \\ > = & \sum_{i=1}^{n} x_{i} \sum_{j=1}^{n} A_{ij} y_{j} \\ > = & \sum_{i,j=1}^{n} x_{i}A_{ij}y_{j}. > \end{align}$ > > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```